只是为了学习sql,我想用一个简单的父子建立一个层次结构.就像堆栈溢出徽章一样(父:问题徽章,儿童:利他主义者).
这是我的sql:
SELECT *
FROM (`badge_types`)
LEFT JOIN `badges` ON `badges`.`badge_type` = `badge_types`.`badge_type_id`
这就是我得到的:
(
[0] => stdClass Object
(
[badge_type_id] => 2
[badge_type_title] => Participation Badges
[badge_type_description] => Badges earning by participating in varIoUs areas of the site.
[badge_type_order] => 2
[badge_id] => 1
[badge_name] => Autobiographer
[badge_level] => 3
[badge_requirement] => Completed all user profile fields
[badge_type] => 2
[badge_order] => 1
[badge_sites] => 0
)
[1] => stdClass Object
(
[badge_type_id] => 1
[badge_type_title] => Experience Badges
[badge_type_description] => Badges earned by amount of experience gain throughout the site.
[badge_type_order] => 1
[badge_id] => 2
[badge_name] => Apprentice
[badge_level] => 3
[badge_requirement] => Achieved 500 experience
[badge_type] => 1
[badge_order] => 1
[badge_sites] => 0
)
)
我怎么能把它变成:
array(
array(
[badge_type_id] => 2
[badge_type_title] => Participation Badges
[badge_type_description] => Badges earning by participating in varIoUs areas of the site.
[badge_type_order] => 2
[badges] => array(
array(
[badge_id] => 1
[badge_name] => Autobiographer
[badge_level] => 3
[badge_requirement] => Completed all user profile fields
[badge_type] => 2
[badge_order] => 1
[badge_sites] => 0
),
array(
[badge_id] => 2
[badge_name] => Example 2
[badge_level] => 3
[badge_requirement] => blah bla
[badge_type] => 2
[badge_order] => 1
[badge_sites] => 0
)
)
),
array(
[badge_type_id] => 1
[badge_type_title] => Experience Badges
[badge_type_description] => Badges earned by amount of experience gain throughout the site.
[badge_type_order] => 1
[badges] => array(
array(
[badge_id] => 2
[badge_name] => Apprentice
[badge_level] => 3
[badge_requirement] => Achieved 500 experience
[badge_type] => 1
[badge_order] => 1
[badge_sites] => 0
),
array(
[badge_id] => 2
[badge_name] => Example 2
[badge_level] => 3
[badge_requirement] => Achieved 1000 experience
[badge_type] => 1
[badge_order] => 1
[badge_sites] => 0
)
)
)
)
我可以使用多个MySQL查询来完成它,但理想情况下我只想使用一个查询,如果可能的话?
解决方法:
使用SQL查询无法实现这一点,因为(由于关系模型性质)SQL查询将始终返回“平面”结果集而不进行任何嵌套(罕见的例外是供应商sql扩展,目标是生成XML输出或类似,但MysqL具有没有这样的扩展).
如果要从sql结果集中获取嵌套数组,则必须使用PHP代码对其进行后处理.可以将此代码组织为像对数组进行分组循环(使用sql预分类),每次“内部”表的键更改时启动新组.这可以通过PHP以相当通用的方式完成,因此您可以编写一个这样的函数来对后处理许多SQL查询(给它适当的参数).
function groupnest( $data, $groupkey, $nestname, $innerkey ) {
$outer0 = array();
$group = array(); $nested = array();
foreach( $data as $row ) {
$outer = array();
while( list($k,$v) = each($row) ) {
if( $k==$innerkey ) break;
$outer[$k] = $v;
}
$inner = array( $innerkey => $v );
while( list($k,$v) = each($row) ) {
if( $k==$innerkey ) break;
$inner[$k] = $v;
}
if( count($outer0) and $outer[$groupkey]!=$outer0[$groupkey] ) {
$outer0[$nestname] = $group;
$nested[] = $outer0;
$group = array();
}
$outer0 = $outer;
$group[] = $inner;
}
$outer[$nestname] = $group;
$nested[] = $outer;
return $nested;
}
data是要嵌套的数组(sql结果集),
groupkey是“外部”实体主键的列名,
innerkey是“内部”实体主键的列名.
在结果集中,“外部”实体的所有列必须在$innerkey列之前,并且“内部”实体的所有列必须遵循它.
要正确分组,结果集必须首先按来自“外部”实体的唯一表达式排序,例如按badge_type_title,badge_type_id,….的顺序排序.后面的字段将按顺序定义“内部”组内的排序.
要嵌套3个或更多实体的连接,您可以多次使用此功能(从内到外折叠“)
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